The null and alternative hypotheses are:
H0:μd≤0H0:μd≤0
H1:μd>0H1:μd>0
The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.
Day | ||||
1 | 2 | 3 | 4 | |
Day shift | 11 | 10 | 14 | 19 |
Afternoon shift | 10 | 9 | 14 | 16 |
At the 0.01 significance level, can we conclude there are more defects produced on the Afternoon shift?
a. State the decision rule. (Round the final answer to 3 decimal places.)
H_{0} is (rejected or not rejected?) if t > _______
b. Compute the value of the test statistic. (Round the final answer to 3 decimal places.)
Test statistic: ___________
c. What is your decision regarding the null hypothesis?
H_{0} should (be rejected or not be rejected?)
d. Determine the p-value. (Round the final answer to 4 decimal places.)
The p-value is ______
a.
Let d be the difference between defects produced on the Afternoon shift and defects produced on the Day shift.
The null and alternative hypotheses are:
H0: μd ≤ 0
H1: μd > 0
Degree of freedom = 4-1 = 3
Critical value of t at df = 3 and 0.01 significance level is 4.541
H0 is rejected if t > 4.541
b.
d = 10-11 , 9-10, 14-14, 16 - 19
= -1, -1, 0, -3
Sample mean, = (-1 - 1 + 0 - 3)/4 = -1.25
Sample standard deviation, s = sqrt[((-1 + 1.25)^2 + (-1 + 1.25)^2 + (0 + 1.25)^2 + (-3 -+1.25)^2 )/(4-1)] = 1.258306
Standard error of mean difference, SE = s / = 1.258306 / = 0.629153
Test statistic, SE = / SE = -1.25 / 0.629153 = -1.987
c.
H0 should not be rejected as t < 4.541
d.
P-value = P(t > -1.987) = 0.9295
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